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\title{Midpoint lenses in segment spaces}
\date{}
\begin{document}
\maketitle
\begin{abstract}
For a real segment-forest dual with unbounded finite component heights
and the real infinite-height coordinate predual, we bound the tail of
an arbitrary displacement in a symmetric lens by $2\sqrt{R^2-\|x\|^2}$.
Here $R$ is the lens radius and $x$ is the center, which we take finitely
supported. Both given norms are asymptotically midpoint uniformly convex,
and neither space admits an equivalent asymptotically uniformly convex
norm. The finite-height forest dual is reflexive.
\end{abstract}

\section{What a midpoint lens measures}\label{sec:intro}

\subsection{Norms and Banach spaces}
A \emph{real normed vector space} is a set $Y$ whose elements can be
added and multiplied by real scalars, together with a \emph{norm}
$x\mapsto\norm{x}$ obeying
\[
 \norm{x}=0 \Longleftrightarrow x=0,\qquad
 \norm{\lambda x}=|\lambda|\norm{x},\qquad
 \norm{x+y}\le\norm{x}+\norm{y}.
\]
Every scalar in this paper is real. The model example is $\R^n$ with
the Euclidean norm
$\norm{(x_1,\dots,x_n)}=\bigl(x_1^2+\dots+x_n^2\bigr)^{1/2}$.
A \emph{Banach space} is a real normed vector space in which every
Cauchy sequence has a limit inside $Y$: no holes, nothing missing.
All spaces in this paper are real Banach spaces.

For a scalar $R>0$, the \emph{closed ball of radius $R$} is
$RB=\{y\in Y:\norm{y}\le R\}$, and the \emph{unit sphere} is
$S=\{y:\norm{y}=1\}$. Balls are convex: if $x,y\in RB$ and
$0\le t\le1$, then $tx+(1-t)y\in RB$.

\subsection{The lens}
A \emph{midpoint lens} is what remains of the ball when we impose two
symmetric constraints at once. Pick a center $x\in Y$ and a radius
$R>0$. The lens at $(x,R)$ is
\begin{equation}\label{eq:lens}
 \mathcal L(x,R) = \bigl\{y\in Y :
                    \norm{x+y}\le R,
                    \ \norm{x-y}\le R\bigr\}.
\end{equation}
The vector $x+y$ is ``$x$ pushed up by $y$'', and $x-y$ is ``$x$
pushed down by $y$''. The lens consists of displacements that keep
both pushed vectors inside the ball of radius $R$. In the plane, with
$Y=\R^2$, the lens is the intersection of two closed disks of radius
$R$, one centered at $x$ and one centered at $-x$.

How thick can a lens be near the boundary? If $x$ already sits inside
$RB$ but near the sphere, and $R$ is only slightly larger, the lens is
a thin shell around $0$. Every question in this paper has the shape:
\emph{how small is $y$ as a function of the squared-radius deficit
$R^2-\norm{x}^2$?}

\subsection{The trivial case: Hilbert space}
A \emph{Hilbert space} is a Banach space whose norm comes from an
inner product, as in $\R^n$ with the Euclidean norm. The
\emph{parallelogram identity} says
\begin{equation}\label{eq:parallelogram}
 \norm{x+y}^2+\norm{x-y}^2 = 2\norm{x}^2 + 2\norm{y}^2
 \qquad(x,y\in Y).
\end{equation}
Suppose $y\in\mathcal L(x,R)$. Adding the two inequalities
$\norm{x+y}^2\le R^2$ and $\norm{x-y}^2\le R^2$, substituting in
\eqref{eq:parallelogram}, and dividing by two gives
\[
 \norm{x}^2+\norm{y}^2\le R^2,\qquad
 \norm{y}^2\le R^2-\norm{x}^2.
\]
In Hilbert space the lens~\eqref{eq:lens} is a ball of radius
$\sqrt{R^2-\norm{x}^2}$, and every displacement is that small. The
quantity $R^2-\norm{x}^2$ is the \emph{squared-radius deficit} of the
center: how far the square of the center length sits below the square
of the lens radius.

In our spaces the situation changes because the norm is not Euclidean:
the parallelogram identity fails, and the pair of lens inequalities no
longer pins down $\norm{y}$ exactly. A bound of the same shape survives
in projection, however, once we project away a finite coordinate head.

\subsection{What this paper proves}
Two closely related spaces appear throughout, and the same main
inequality holds for both.

The first is a real Banach space $X=J^*$, the dual of a specific
tree-number space $J$ whose norm is built from sums over disjoint
segments. It is \emph{reflexive} ($X^{**}=X$: the double dual
recovers the original space) and \emph{separable} (a countable dense
subset exists). The model to keep in mind is a Hilbert space $\ell_2$
renormed so that long chains are more expensive than spread-out mass.

The second is
$B_\infty \subsetneq J_\infty^*$, the closed coordinate span inside
the dual of the same construction on a single infinite rooted tree
$T_\infty=\N^{<\omega}$ of finite words. Here the closure of the
coordinate span is strictly smaller than the full dual; this is the
point of the distinction.

For each space, $H$ denotes a \emph{finite ancestral set}: a finite
set of vertices that contains every ancestor of each of its members.
$P_H$ projects to the coordinates indexed by $H$, and $Q_H=I-P_H$ is
the complementary projection.

\begin{theorem}[Main lens bound]\label{thm:main}
Let $Y$ be $X$ or $B_\infty$. Let $H$ be a finite ancestral set, and let
$x,y\in Y$ satisfy $P_Hx=x$. For every $R\ge0$ such that
$\norm{x+y}\le R$ and $\norm{x-y}\le R$,
\begin{equation}\label{eq:main-bound}
 \norm{Q_Hy}\le 2\sqrt{R^2-\norm{x}^2}.
\end{equation}
\end{theorem}

Heads and radii are chosen for each $x$ and each $R$ once; a single $H$
then controls \emph{all} displacements $y$ of the lens at $(x,R)$
simultaneously. That uniformity is the central feature. It makes the
estimate compatible with the cofinite infimum that appears in the
midpoint modulus, defined shortly.

\subsection{Midpoint moduli}
For $t>0$, an infinite dimensional Banach space $Y$, and the family of
closed linear subspaces of finite codimension, the
\emph{averaged asymptotic midpoint modulus} is
\begin{equation}\label{eq:avg-modulus}
 \widehat\delta_Y(t)
 = \inf_{\substack{x\in Y\\ \norm{x}=1}}
   \sup_{F\subseteq Y \text{ closed, } \dim Y/F<\infty}
   \inf_{\substack{y\in F\\\norm{y}=1}}
   \left(
    \frac{\norm{x+ty}+\norm{x-ty}}{2}-1
   \right).
\end{equation}
The inner expression is the excess of the average pair-length over $1$.
The space $Y$ is \emph{asymptotically midpoint uniformly convex}
(or AMUC) when $\widehat\delta_Y(t)>0$ for every $t>0$. Intuition:
no matter which unit directions you probe, a fixed proportional step
$t$ cannot leave both $x+ty$ and $x-ty$ on the spherical cap. Finite-
codimensional head coordinates can be killed, but the remaining tail
dilutes the average. So the excess is uniformly positive.

Similarly define the \emph{maximum-based} modulus
\begin{equation}\label{eq:max-modulus}
 \overline\theta_Y(t)
 = \inf_{\norm{x}=1}
   \sup_F
   \inf_{\norm{y}=1\ ,\ y\in F}
   \max\{\norm{x+ty},\norm{x-ty}\}-1.
\end{equation}
The two moduli are qualitatively equivalent:
$\frac12\overline\theta_Y(t)\le\widehat\delta_Y(t)\le\overline\theta_Y(t)$.
The right inequality is pointwise (average $\le$ maximum). For the left,
fix a unit center $x$ and a norming functional $f\in Y^*$ with
$\norm{f}=1$ and $f(x)=1$. Intersect any cofinite $F$ with $\ker f$.
Both endpoints then satisfy $f(x\pm y)\ge1-\norm{y}$, so $\norm{x\pm y}\ge1$,
and an excess $a$ in the maximum becomes at least $a/2$ in the average.

The two conventions agree qualitatively. They differ numerically: a
bound derived for the average typically has a cleaner constant than the
same bound converted to the maximum.

\subsection{Why this is interesting}
Hilbert space already has $\widehat\delta_{\mathbb H}(t)
=\bigl(\sqrt{1+t^2/4}-1\bigr)/2$, obtained by solving
\eqref{eq:parallelogram}. The question is whether AMUC survives when
the norm is fundamentally non-Hilbert.

Baudier~\cite{Baudier2026} showed that the Kadets--Werner modification
of the Bourgain--Rosenthal space is AMUC without any equivalent AUC
norm. Earlier, Girardi~\cite{Girardi2001} proved asymptotic uniform
convexity of the dual of the binary James tree space. The spaces in
this paper use countably branching trees of unbounded finite heights
(or a single infinite-height tree), and the main result gives:
\begin{itemize}
\item the twin spaces $X$ and $B_\infty$ are AMUC;
\item neither carries an equivalent AUC norm.
\end{itemize}
The whole argument is proved from scratch below, beginning with the
vocabulary in Section~\ref{sec:vocab}.

Perreau~\cite{Perreau2021} asked whether the canonical predual of the
countably branching James tree space is AMUC. The space $B_\infty$
here is closely related but not canonically identical to any standard
construction; this paper does not claim to settle Perreau's question in
its original wording.

\subsection{Reading guide}
Section~\ref{sec:vocab} collects the standard words (norm, dual,
weak nullity) that appear everywhere. Section~\ref{sec:spaces}
builds the two tree spaces and their primal counterparts.
Section~\ref{sec:projections} extracts the coordinate facts: projections
are contractive, and gates decompose tails. Section~\ref{sec:atoms}
linearizes finite head balls into atoms. Section~\ref{sec:crossing}
proves the cancellation lemma that controls opposing signs across the
head boundary. Section~\ref{sec:main-theorem} assembles the lens bound
of Theorem~\ref{thm:main}. Section~\ref{sec:moduli} converts it to the
averaged modulus. Section~\ref{sec:obstruction} proves the obstruction
to an equivalent AUC norm. Section~\ref{sec:variants} catalogues the
alternative proofs from the longer manuscript, with their numerical
constants, so a reader can compare methods without reading every page.

\section{Vocabulary from analysis}\label{sec:vocab}

\subsection{Duals and weak nullity}
Let $Y$ be a real Banach space. A \emph{bounded linear functional} on
$Y$ is a linear map $f:Y\to\R$ whose operator norm
\[
 \norm{f}=\sup_{\norm{y}\le1}|f(y)|
\]
is finite. The \emph{dual space} $Y^*$ is the Banach space of all such
$f$. A functional $f\in Y^*$ is \emph{norming} for a nonzero $x\in Y$
when $\norm{f}=1$ and $f(x)=\norm{x}$. The Hahn--Banach theorem
guarantees a norming functional for every $x\neq0$; uniqueness
generally fails and is not needed.

Each $y\in Y$ defines a functional on $Y^*$ by evaluation
$f\mapsto f(y)$. This is a canonical linear isometric injection
$Y\hookrightarrow Y^{**}$, and $Y$ is \emph{reflexive} when this map
is onto. Reflexivity allows us to recover the norm of any $y$ by
testing against functionals $f\in Y^*$ with $\norm{f}\le1$.

A \emph{cofinite subspace} of $Y$ is a closed linear subspace of finite
codimension. Equivalently, $F$ is the set of common solutions of
finitely many bounded linear equations $f_1(y)=\dots=f_k(y)=0$. This
is the only kind of restriction we ever impose: forbid a finite set of
linear observations, and no more.

A sequence $(y_j)\subset Y$ is \emph{weakly null} if $f(y_j)\to0$ for
every $f\in Y^*$. A weak null sequence can still avoid the origin in
norm; it merely leaves the stage through every individual observation.
For example, if $(e_j)$ is an orthonormal basis of a Hilbert space,
then $(e_j)$ is weakly null even though $\norm{e_j}\equiv1$.

\subsection{$\ell_2$ direct sums}
For real Banach spaces $A$ and $B$, the $\ell_2$ direct sum is
\[
 A\oplus_2 B = \{(a,b):a\in A,\ b\in B\},\qquad
 \norm{(a,b)} = \bigl(\norm{a}^2+\norm{b}^2\bigr)^{1/2}.
\]
For a countable family $(A_i)_{i\in I}$,
\[
 \bigl(\bigoplus_{i\in I}A_i\bigr)_{\ell_2}
 = \Bigl\{(a_i)_{i\in I} : \textstyle\sum_i\norm{a_i}^2<\infty\Bigr\},
 \quad
 \norm{(a_i)} = \Bigl(\sum_i\norm{a_i}^2\Bigr)^{1/2}.
\]
Duality passes through:
$(\bigoplus_i A_i)^*=(\bigoplus_i A_i^*)_{\ell_2}$, and reflexivity of
the summands passes up to reflexivity of the sum.

\subsection{The one-sided AUC modulus}
For a norm $N$ on a Banach space $Z$, define the \emph{one-sided
asymptotic modulus}
\begin{equation}\label{eq:one-sided-modulus}
 \overline\delta_N(t)
 = \inf_{\substack{z\in Z\\ N(z)=1}}
   \sup_{\substack{F\subseteq Z\\ \text{cofinite}}}
   \inf_{\substack{w\in F\\ N(w)=1}}
   \bigl(N(z+tw)-1\bigr),\qquad t>0.
\end{equation}
The infimum over the unit center $z$ and cofinite escape $F$ asks for
the worst-case growth of a single pushed-up vector $z+tw$. A norm $N$
is \emph{asymptotically uniformly convex} (AUC) when
$\overline\delta_N(t)>0$ for every $t>0$. This is strictly stronger
than AMUC: only $z+tw$ is tested, never $z-tw$. Section~\ref{sec:obstruction}
exhibits a family of unit root-path vectors and unit child increments,
showing that no equivalent renorming of $X$ or $B_\infty$ can be AUC.

\subsection{Summary of notation}
Uppercase Latin letters $x,y,z,u,v,w$ denote vectors; lowercase
$t,s,\lambda,\mu$ denote real scalars; $H$ stands for a finite
ancestral head; $P_H$ projects to the coordinates $H$; $Q_H$ is the
complement; $G_H$ is the set of minimal excluded vertices (the gates);
$J$ and $J_\infty$ are the primal segment spaces; $X=J^*$ and
$B_\infty\subseteq J_\infty^*$ are the two target spaces; $\rho$ is
the primal segment test norm; and $\norm{\cdot}$ denotes the target
norm.

\section{Tree spaces and their segment norms}\label{sec:spaces}

\subsection{Rooted trees of words}
For an integer $h\ge1$, let
\[
 T_h=\N^{\le h}
 =\{\text{finite words in }1,2,3,\dots\text{ of length }\le h\}
\]
be the rooted tree whose root is the empty word $o$ and whose edges
join each word $s$ to $s^\frown j=sj$. A \emph{vertex} is a word;
its \emph{length} is the number of letters in it. The \emph{ancestors}
of a word $v$ are the prefixes of $v$, including $v$ itself; every
vertex has countably many \emph{children} $s^\frown1,s^\frown2,\dots$,
and the deepest vertices of $T_h$ have length $h$. The tree $T_h$ has
\emph{height} $h$: no branch is longer than $h$ edges. Let
$T_\infty=\N^{<\omega}$ be the same tree without a height bound.

A \emph{finite forest} is a disjoint union of components, one for each
height:
\[
 \mathcal F=\coprod_{h\ge1}(\{h\}\times T_h).
\]
Vertex $(h,s)$ of $\mathcal F$ lives in component $h$ and sits at word
$s$; components of higher $h$ are taller. So $\mathcal F$ has
countably many components and the component heights are unbounded.

Our first target space is attached to $\mathcal F$; our second is
attached to $T_\infty$ alone.

\subsection{Segments and disjointness}
A \emph{segment} of a rooted tree is a set
$[a,b]=\{v : a\preceq v\preceq b\}$ for an ancestor $a$ and a
descendant $b$ of $a$, both inclusive. Equivalently, a segment is a
vertical chain in the tree with a top and a bottom. In $T_1$, every
segment is a singleton or a two-vertex path; in a very tall component,
a segment can be a long chain. Segments are confined to one component
of $\mathcal F$; they never cross from $\{h\}\times T_h$ to
$\{h'\}\times T_{h'}$ when $h\ne h'$.

Two segments are \emph{disjoint} if the sets of vertices are disjoint.
Because each segment is a chain, disjoint segments typically live in
incomparable branches of the tree, or in different components.

\subsection{The segment norm}
Let $u:\mathcal F\to\R$ be a function of finite support (zero for all
but finitely many vertices). For a segment $S$, the \emph{segment sum}
is
\[
 u(S)=\sum_{v\in S}u(v).
\]
The \emph{segment norm} of $u$ is
\begin{equation}\label{eq:segment-norm}
 \rho(u) =
 \sup_{\substack{S_1,\dots,S_m\text{ disjoint segments}\\m\ge1}}
 \Bigl(\sum_{i=1}^m u(S_i)^2\Bigr)^{1/2}.
\end{equation}
The supremum is over all finite families of pairwise disjoint segments.
Using single-vertex segments $S=\{v\}$, we get
$\max_v|u(v)|\le\rho(u)$; on the other hand, decomposing a family of
disjoint segments into its individual vertices shows
$\rho(u)\le\sum_v|u(v)|$. So $\rho$ is squeezed between the
$\ell_\infty$ and $\ell_1$ norms: it is a genuine norm.

\emph{Example: a solid chain.} Suppose $u$ is supported on one long
segment $S$ and is identically one on it. Any family of disjoint
subsegments of $S$ has total length at most $|S|$, and the sum of
their squares is largest when the family is $S$ alone, so
$\rho(u)=|S|$, the length of $S$. A long unbroken chain is therefore
very costly under $\rho$.

\emph{Example: an antichain.} Suppose $u$ is supported on an antichain
$A$ (a set of vertices no two of which are comparable). Every segment
meets $A$ in at most one vertex, so any disjoint family picks out at
most one $u(v)$ from $A$ per segment. Taking all singletons gives
$\rho(u)=\bigl(\sum_{v\in A}u(v)^2\bigr)^{1/2}$. On an antichain,
the segment norm collapses to the ordinary Euclidean norm. This is why
the space below is Hilbert-like when it is spread out over many
independent branches.

\subsection{The primal space $J$ and its dual $X$}
Take the completion of the finitely supported functions under $\rho$;
call the result $J$. Its dual
\[
 X=J^* = \{\text{bounded linear functionals on }J\}
\]
is our first target space. On $T_\infty$, the same construction defines
the primal space $J_\infty$. The two cases use the same notation when
there is no ambiguity; we simply say ``$J$'' for either.

For future use, note the coefficient inequality
\begin{equation}\label{eq:coefficient}
 \Bigl\|\sum_i a_i d_{S_i}\Bigr\|_{J^*}
 \le \Bigl(\sum_i a_i^2\Bigr)^{1/2}
 \qquad(S_i\text{ pairwise disjoint}),
\end{equation}
where $d_S=\sum_{v\in S}e_v$ is the \emph{segment functional}
associated with $S$. This is the immediate consequence of
\eqref{eq:segment-norm}: for $u$ with $\rho(u)\le1$,
\[
 \Bigl|\sum_i a_i u(S_i)\Bigr|
 \le \Bigl(\sum_i a_i^2\Bigr)^{1/2}\Bigl(\sum_i u(S_i)^2\Bigr)^{1/2}
 \le \Bigl(\sum_i a_i^2\Bigr)^{1/2}.
\]

\subsection{The infinite-height coordinate predual $B_\infty$}
On $T_\infty$, define
\[
 B_\infty
 = \overline{\operatorname{span}}\{e_v^* : v\in T_\infty\}
 \subseteq J_\infty^*,
\]
where the closure is in $J_\infty^*$ and $e_v^*:u\mapsto u(v)$ is the
evaluation functional at vertex $v$. Each $e_v^*$ has norm one
(testing against $e_v$). This closed coordinate span is strictly
smaller than $J_\infty^*$. Indeed fix an infinite branch
$\beta\subseteq T_\infty$; the finitely supported sum
$f_\beta(u)=\sum_{v\in\beta}u(v)$ extends by continuity to a norm-one
element of $J_\infty^*$, and $f_\beta$ has distance $1$ from
$B_\infty$: for any $y\in B_\infty$ and any vertex $v\in\beta$ beyond
$y$'s support, $\norm{f_\beta-y}\ge|f_\beta(e_v)-y(e_v)|=1$.

So $B_\infty$ differs from $X$ in two ways: its primal is the single
infinite tree, and its target is squeezed to a coordinate span inside
the full dual.

\subsection{Basic bounds and standing assumptions}
For every $x\in X$ or $x\in B_\infty$ and every finitely supported
$u$,
\begin{equation}\label{eq:basic-bounds}
 \norm{x}_\infty\le\norm{x}\le\norm{x}_2,\qquad
 \rho(u)_\infty\le\rho(u)\le\rho(u)_1.
\end{equation}
A single coordinate is never larger than the norm, and the Euclidean
coefficient norm dominates it. On the primal side $\rho$ sits between
the $\ell_\infty$ and $\ell_1$ norms. On an antichain all four
comparisons are equalities.

\emph{Standing assumptions.} From now on:
\begin{itemize}
\item ``Either space'' means $X=J^*$ or $B_\infty$; both are real.
\item ``$J$'' means $J$ or $J_\infty$.
\item ``A vertex $v$'' is an element of $\mathcal F$ or of $T_\infty$.
\item ``A finite ancestral set $H$'' is a finite set of vertices that
      contains every ancestor of each of its members.
\end{itemize}
Every subsequent theorem explicitly names which of the two spaces it
concerns, because some of the proofs use reflexivity of $X$ whereas
$B_\infty$ requires the coordinate-span formulation.

\section{Coordinate structure of the forest}\label{sec:projections}

\subsection{Ancestral heads and projections}
A finite set $H$ of vertices is \emph{ancestral} if whenever $v\in H$
and $u\preceq v$, one also has $u\in H$. Intuition: to include a
vertex, one must include all of its ancestors up to the root. On the
forest, roots are the bottom vertices of \emph{components}; on
$T_\infty$, the root is the empty word $o$.

For a finite ancestral set $H$, define
\[
 P_H:Y\to Y,\quad P_Hx=\sum_{v\in H}x(e_v)e_v^*,\qquad
 Q_H=I-P_H.
\]
The projection $P_H$ records the coordinates at $H$; $Q_H$ discards
them. We call $Q_Hy$ the \emph{tail} of $y$ outside $H$.

\begin{lemma}[Contraction]\label{lem:projections}
On either space $Y=X$ or $Y=B_\infty$, and for any finite ancestral
$H$, both $P_H$ and $Q_H$ are bounded linear operators of norm at
most one.
\end{lemma}

\begin{proof}
A segment of the tree meets $H$ in a (possibly empty) prefix and
$\mathcal F\setminus H$ in a (possibly empty) suffix. If
$S_1,\dots,S_m$ are disjoint segments, then their intersections
$S_i\cap H$ remain disjoint, and $u(S_i\cap H)$ is the segment sum of
$P_Hu$ on that intersection. Applying the defining
supremum~\eqref{eq:segment-norm} to $P_Hu$ therefore gives
$\rho(P_Hu)\le\rho(u)$. Extending by density to $J$ and passing to
adjoints on the duals $X$ and $B_\infty$ proves the bound for $P_H$.
For $Q_H$, the same argument applies to the complements
$S_i\setminus H$, which are also disjoint segments (possibly empty).
\end{proof}

\begin{lemma}[Coordinate density]\label{lem:density}
Finitely supported vectors are norm-dense in $X$ and in $B_\infty$.
Along any increasing sequence of finite ancestral sets
$H_1\subseteq H_2\subseteq\dots$ whose union is $\mathcal F$ or
$T_\infty$, one has $P_{H_n}x\to x$ in norm for every $x$.
\end{lemma}

\begin{proof}
For $B_\infty$ the space is by definition the norm closure of the
finitely supported coordinate span, so finite vectors are dense by
construction.

For $X=J^*$, let $Z\subseteq X$ be the norm closure of the finitely
supported coordinate functionals. If $Z\ne X$, Hahn--Banach gives a
nonzero $u\in X^*$ that vanishes on $Z$. By reflexivity of $X$
(Lemma~\ref{lem:reflexive}), $X^*=J$, so $u\in J$; in particular
$u(e_v^*)=u(v)=0$ for every vertex $v$. Finitely supported functions
are dense in $J$, so $u=0$, a contradiction. Hence $Z=X$.

For the projection statement fix $x$ and $\eps>0$. By the density just
proved, choose a finitely supported $y$ with $\norm{x-y}<\eps$. Once
$H_n$ contains the support of $y$ one has $P_{H_n}y=y$, so
contractivity of $P_{H_n}$ gives
\[
 \norm{P_{H_n}x-x}\le\norm{P_{H_n}(x-y)}+\norm{y-x}\le2\eps.
\]
\end{proof}

\subsection{Gates and the cone decomposition}
The minimal vertices of $\mathcal F\setminus H$ are \emph{gates} of
$H$; write $G_H$ for the set. Because $H$ is ancestral, distinct gates
are pairwise incomparable. The descendant subtrees
\[
 U_g=\{v\in\mathcal F : g\preceq v\}\qquad(g\in G_H)
\]
partition $\mathcal F\setminus H$, and each $U_g$ is itself a
countably branching tree rooted at $g$. Any segment that crosses the
boundary between $H$ and its complement enters the complement at a
unique gate.

\begin{lemma}[Cone decomposition]\label{lem:cones}
For a finite ancestral $H$ and $x\in Y$,
\begin{equation}\label{eq:cone-sum}
 \norm{Q_Hx}^2=\sum_{g\in G_H}\norm{x\mathbf 1_{U_g}}^2.
\end{equation}
Also, the \emph{gate-coordinate map}
$x\mapsto(x(e_g))_{g\in G_H}$ is a contraction from $Y$ to
$\ell_2(G_H)$.
\end{lemma}

\begin{proof}
For the equality, note first that on the primal side every segment
of the complement lies wholly in a single $U_g$. Any disjoint family
$S_1,\dots,S_m\subset\mathcal F\setminus H$ can be partitioned into
groups indexed by $g$, and the contributions to $\rho^2$ split:
$\rho(u)^2=\sum_{g}\rho(u\mathbf 1_{U_g})^2$ for finitely supported
$u$ in $\mathcal F\setminus H$. Extend by density to $J$ and dualise
to get the sum identity on $X$ and on $B_\infty$.

For the contraction, let $A\subseteq G_H$ be finite and let
$a=(a_g)_{g\in A}$ be real coefficients. The vector
$v_A=\sum_{g\in A}a_g e_g\in J$ is finitely supported on an antichain
$A$; since every segment meets $A$ in at most one vertex,
$\rho(v_A)=\bigl(\sum_{g\in A}a_g^2\bigr)^{1/2}$. Then
\[
 \Bigl|\sum_{g\in A}a_g x(e_g)\Bigr|
 =|x(v_A)|\le\norm{x}\,\rho(v_A)
 =\norm{x}\Bigl(\sum_{g\in A}a_g^2\Bigr)^{1/2}.
\]
Taking the supremum over $a$ proves the contraction.
\end{proof}

\subsection{Reflexivity and separability of $X$}
\begin{lemma}[Reflexivity and separability]\label{lem:reflexive}
The space $X=J^*$ is reflexive and separable.
\end{lemma}

\begin{proof}
Fix $h\ge1$ and consider the component
$J_h=\overline{c_{00}(T_h)}^{\rho}$. Every segment in $T_h$ has at
most $h+1$ vertices, so the elementary estimate
\[
 \norm{u}_2\le\rho_h(u)\le\sqrt{h+1}\,\norm{u}_2
\]
holds. Hence $J_h$ is Hilbert-isomorphic as a topological vector
space, and consequently separable and reflexive. Combining the
components in $\ell_2$:
$J=(\bigoplus_{h\ge1}J_h)_{\ell_2}$. The $\ell_2$ sum of separable
reflexive spaces is separable and reflexive, so $J$ is both. Its dual
$X=J^*$ is reflexive by the same argument applied to the reverse side,
and separable because the dual of a separable Banach space is
separable in the weak-$*$ topology; here reflexivity upgrades that
to norm separability. (A direct proof also works: coordinate
functionals with rational coefficient and finite support form a
countable dense set.)
\end{proof}

\subsection{Norming tests attached to finite heads}
\begin{lemma}[Norming test in a finite head]\label{lem:norming}
Let $H$ be finite ancestral, and let $0\ne x=P_Hx\in Y$. There exists
$p=P_Hp\in J$ with $\rho(p)=1$ and $x(p)=\norm{x}$.
\end{lemma}

\begin{proof}
The space $P_HY$ is finite dimensional, so the closed unit ball of
$P_HJ$ is compact. The functional $v\mapsto x(v)$, defined for
$v\in P_HJ$ with $\rho(v)\le1$, attains a maximum. Because $x$
vanishes on the $J$-annihilator of $P_HJ$ (the functionals supported
on $\mathcal F\setminus H$), any maximizer may be taken supported on
$H$. Rescale the maximizer so that $\rho(p)=1$.
\end{proof}

\section{Atoms: convex models of a finite head}\label{sec:atoms}

\subsection{Segment functionals}
For a segment $S$ of the tree, define the \emph{segment functional}
\[
 d_S = \sum_{v\in S}e_v\in J,\qquad
 d_S(u) = u(S) = \sum_{v\in S}u(v).
\]
The norm of $d_S$ in $J^*$ is exactly one. Indeed, by the definition
of $\rho$, the single-element family $\{S\}$ gives
$\rho(u)\ge|u(S)|$; hence $|d_S(u)|\le\rho(u)$ for all $u$. Taking
$u=\mathbf 1_S$ (the indicator of $S$), we have $u(S)=|S|$ and
$\rho(u)=|S|$ (see the ``solid chain'' example above), so
$|d_S(u)|/\rho(u)=1$. Therefore $\norm{d_S}=1$.

\subsection{Atoms and their coefficient bound}
An \emph{atom} is a functional of the form
\begin{equation}\label{eq:atom}
 A = \sum_{i=1}^m a_i d_{S_i},
 \qquad S_1,\dots,S_m\text{ pairwise disjoint},
 \qquad \sum_{i=1}^m a_i^2\le1.
\end{equation}
The segment family $(S_i)$ and coefficients $(a_i)$ are part of the
data; the same point of $J^*$ can have several atomic representations.
By the Cauchy--Schwarz argument sketched under~\eqref{eq:coefficient},
an atom has norm at most one:
\begin{equation}\label{eq:atom-norm}
 |A(u)|
 =\Bigl|\sum_i a_i u(S_i)\Bigr|
 \le\Bigl(\sum_i a_i^2\Bigr)^{1/2}
     \Bigl(\sum_i u(S_i)^2\Bigr)^{1/2}
 \le 1\cdot\rho(u).
\end{equation}
The first inequality is Cauchy--Schwarz in $i$; the second uses the
definition of $\rho$ on the disjoint segment family. Atoms are the
building blocks from which the finite-stage unit balls are recovered.

\subsection{The finite atomic ball}
\begin{lemma}[Finite atomic ball]\label{lem:finite-hull}
Let $H$ be a finite ancestral set. The closed unit ball of $P_HY$
equals the convex hull of atoms supported in $H$. Every element of
that convex hull is a convex combination of at most $|H|+1$ atoms
with segments in $H$.
\end{lemma}

\begin{proof}
There are finitely many families of disjoint segments in the finite set
$H$. For each such family $(S_1,\dots,S_m)$, the corresponding atoms
form the image under a linear map of the Euclidean ball
$\{a\in\R^m : \sum a_i^2\le1\}$, which is compact. The union over all
families is compact; its convex hull $C$ is therefore a compact convex
subset of the finite-dimensional space $P_HJ^* = P_HY$.

For a test $u\in P_HJ$ with $\rho(u)\le1$, the \emph{support function}
of $C$ at $u$ computes
\[
 \sup_{A\in C}A(u)
 =\sup_{\text{disjoint families}}\Bigl(\sum_i u(S_i)^2\Bigr)^{1/2}
 =\rho(u).
\]
This is exactly the support function of the closed unit ball of
$P_HY$. In finite dimension, two closed convex sets with the same
support function are equal. Therefore $C$ is the unit ball of
$P_HY$.

The Carathéodory bound: a convex body in a real vector space of
dimension $d$ is the convex hull of any set whose every member is
expressible as a convex combination of at most $d+1$ of the set's
points. Here $P_HY$ has dimension $|H|$ (its coordinates are indexed
by $H$, and $\R$ has dimension one), so every point of the unit ball
is a convex combination of at most $|H|+1$ atoms.
\end{proof}

\begin{corollary}[Completed atomic ball]\label{cor:completed-ball}
The unit ball of $Y$ is the norm closure of the union over finite
ancestral $H$ of the finite-stage atomic balls of
Lemma~\ref{lem:finite-hull}. Equivalently, $B_Y$ is the norm-closed
convex hull of all atoms in $Y$.
\end{corollary}

\begin{proof}
Every atom has norm at most one, so the norm-closed convex hull of
all atoms lies in $B_Y$. Conversely, take any $x\in B_Y$. Fix an
ancestral exhaustion $H_1\subseteq H_2\subseteq\dots$ of $\mathcal F$
(or of $T_\infty$); by density~\eqref{lem:density},
$P_{H_n}x\to x$. Contractivity of $P_H$ gives
$\norm{P_{H_n}x}\le\norm{x}\le1$. By the finite atomic ball,
$P_{H_n}x$ is a convex combination of atoms supported in $H_n$. So
$x$ is a limit of points in the convex hull of all atoms, and
therefore belongs to its norm closure.
\end{proof}

\subsection{Uniqueness of a crossing per atom}
A segment may encounter many vertices of a finite ancestral head $H$,
but only one gate of $H$ (the first vertex that lies outside $H$).
Because an atom's segments are pairwise disjoint, at most one atom
segment crosses the boundary at any fixed gate.

\begin{lemma}[At most one crossing per gate]\label{lem:one-crossing}
Let $H$ be finite ancestral, let $g\in G_H$ be a gate, and let
$A=\sum_i a_i d_{S_i}$ be an atom. There is at most one index $i$
such that $S_i$ meets $H$ and its first exterior vertex is $g$.
\end{lemma}

\begin{proof}
Two such segments would both contain the vertex $g$, contradicting
disjointness. Distinct gates are incomparable, so a single segment
$S_i$ can meet at most one gate: if $g,g'\in G_H$ with $g\prec g'$
and both $g,g'\in S_i$, then $g'$ would be exterior but $g$ interior,
a contradiction with the definition of gate. So each atom segment
crosses at no more than one gate.
\end{proof}

\begin{lemma}[Finite head test]\label{lem:finite-head-norming}
For $x=P_Hx\in Y$ with $x\ne0$, the norming test $p\in J$ of
Lemma~\ref{lem:norming} can be chosen with the following additional
property: the segment sums $p(S_i)$ computed at an atom's segment
family are well-defined for every $S_i$, so an atom can be
\emph{evaluated} on the head test.
\end{lemma}

\begin{proof}
The maximizer $p$ in Lemma~\ref{lem:norming} is finitely supported
inside $H$; segment sums $p(S_i)=\sum_{v\in S_i\cap H}p(v)$ are then
finite for any atom's segment family, even if $S_i$ extends beyond $H$.
\end{proof}

\subsection{The main use of atoms}
The finite atomic ball will be used as follows. In the proof of
Theorem~\ref{thm:main}, the two endpoint vectors $x\pm y$ (scaled to
lie in the unit ball of $Y$) are approximated by finite convex
combinations of atoms. The atom structure lets us track (i) the
\emph{total coefficient energy} $\sum a_i^2\le1$, (ii)
the \emph{segment sums} $p(S_i)$ of a norming head test $p$, and
(iii) the \emph{gate values} $A(e_g)$ crossing from $H$ to its
complement. Section~\ref{sec:crossing} controls the interaction of
gate values of two atoms that leave the head at the same gate with
opposite signs.

\section{Cancellation across a gate}\label{sec:crossing}

\subsection{Crossing sums}
Fix a finite ancestral set $H$ and a segment test $p\in J$ supported in
$H$, with $\rho(p)\le1$. For a disjoint family $\mathcal S$ of segments
define, at each gate $g\in G_H$:
\begin{itemize}
\item the \emph{crossing sum} $b_g(\mathcal S)=p(S)$ if there is a
unique $S\in\mathcal S$ whose head $S\cap H$ has bottom vertex
$\operatorname{parent}(g)$ and whose first exterior vertex is $g$, and
$b_g(\mathcal S)=0$ if no such $S$ exists;
\item the \emph{unused squared-test energy}
\begin{equation}\label{eq:unused-energy}
 L_{\mathcal S}=1-\sum_{S\in\mathcal S}p(S)^2\ge0.
\end{equation}
\end{itemize}
The nonnegativity holds because $\mathcal S$ is a disjoint segment
family and $\rho(p)\le1$, so $\sum_{S\in\mathcal S}p(S)^2\le1$. For an
atom $A$ with segment family $(S_i)$, write $b_g(A)=b_g((S_i))$ and
$L_A=L_{(S_i)}$; by Lemma~\ref{lem:one-crossing} at most one segment
of $A$ crosses any given gate, so $b_g(A)$ is unambiguous.

Since $p$ is supported in $H$, we also have $p(S)=p(S\cap H)$ for any
segment $S$: the tail portion contributes nothing. Consequently
$b_g(\mathcal S)$ depends only on the head portions $S\cap H$ of the
segments in $\mathcal S$.

\subsection{The opposite crossing lemma}
\begin{lemma}[Opposite crossings]\label{lem:opposite}
For any two disjoint segment families $\mathcal S,\mathcal S'$ and any
test $p\in J$ supported on $H$ with $\rho(p)\le1$,
\begin{equation}\label{eq:opposite}
 2\sum_{\substack{g\in G_H\\ b_g(\mathcal S)b_g(\mathcal S')<0}}
    |b_g(\mathcal S)b_g(\mathcal S')|
 \le L_{\mathcal S}+L_{\mathcal S'}.
\end{equation}
\end{lemma}

\begin{proof}
Fix a gate $g\in G_H$ where $b=b_g(\mathcal S)$ and $b'=b_g(\mathcal S')$
have opposite signs. Both crossing segments end their $H$-portion at
the same vertex $v=\operatorname{parent}(g)$. Writing the two heads as
$[a,v]$ and $[a',v]$, we have $a,a'\preceq v$; both $a$ and $a'$ lie on
the same ancestral chain up to $v$, so the heads are nested. They cannot
be equal: if $[a,v]=[a',v]$, then $p([a,v])=p([a',v])$, so $b=b'$, and
$bb'=(b)^2\ge0$, contradicting $bb'<0$. Therefore one head is strictly
larger; without loss $a\prec a'$, so $[a,v]\supsetneq[a',v]$.

Let the crossing segment in $\mathcal S$ be $S_i=[a,w]$. Split it at
$\operatorname{prev}(a')$:
\[
 S_i=[a,w] \rightsquigarrow [a,\operatorname{prev}(a')]
  \quad\text{and}\quad
  [a',w].
\]
Both pieces are nonempty (because $a\prec a'$, and $w\notin H$ by
crossing), are subsets of $S_i$, and remain disjoint from every other
segment of $\mathcal S$. The $p$-sums of the two pieces are
\[
 p([a,\operatorname{prev}(a')])=p([a,v])-p([a',v])=b-b',
\qquad
 p([a',w])=p([a',v])=b',
\]
using $p$ supported in $H$. So the squared $p$-sum changes from $b^2$
to
\[
 (b-b')^2+(b')^2
 =b^2-2bb'+2(b')^2,
\]
an increase of $2(b')^2-2bb'\ge2|bb'|$: the last inequality uses
$bb'<0$, which forces $-2bb'\ge2|bb'|$.

At every opposite-sign gate, split the longer head in whichever family
contains it. Different gates require different segments (Lemma~\ref{lem:one-crossing}),
so each segment of $\mathcal S$ is cut at most once, and each segment
of $\mathcal S'$ at most once. The new family $\tilde{\mathcal S}$
of the modified $\mathcal S$ is still disjoint. Because
$\rho(p)\le1$ and $\tilde{\mathcal S}$ is a disjoint segment family,
\[
 \sum_{S\in\tilde{\mathcal S}}p(S)^2\le1.
\]
Hence the total increase in the $\mathcal S$ squared sum satisfies
\[
 2\sum_{\substack{g:\,b_g(\mathcal S)b_g(\mathcal S')<0\\ \text{head in }\mathcal S\text{ longer}}}
      |b_g(\mathcal S)b_g(\mathcal S')|
 \le L_{\mathcal S}.
\]
The same argument for $\mathcal S'$ gives
\[
 2\sum_{\substack{g:\,b_gb_g'<0\\ \text{head in }\mathcal S'\text{ longer}}}
      |b_gb_g'|
 \le L_{\mathcal S'}.
\]
Adding the two inequalities proves~\eqref{eq:opposite}.
\end{proof}

\subsection{The two-piece picture}
Figure~\ref{fig:opposite-crossing} distinguishes the two original
segments from the two pieces created by the split. On the left, two
crossing segments leave the head at the same vertex $v$ and enter the
complement at the same gate $g$, with opposite $p$-sums $b$ and $b'$.
On the right, the longer head has been split just before the shorter
head starts.

\begin{figure}[tb]
\centering
\begin{tikzpicture}[x=1cm,y=0.75cm,font=\small,
  head/.style={line width=1.4pt,blue!65!black},
  second/.style={line width=1.4pt,red!65!black},
  tail/.style={dashed,line width=1pt},
  vertex/.style={circle,fill=black,inner sep=1.5pt}]
  \node at (1.4,4.6) {Two crossing heads};
  \draw[gray!70,densely dotted] (-0.6,0.4)--(3.1,0.4);
  \node[anchor=west] at (2.6,0.85) {\(H\)};
  \node[anchor=west] at (2.6,-0.05) {\(H^c\)};
  \draw[head] (0,3.8)--(0,0.8);
  \draw[head,tail] (0,0.8)--(0,-0.7);
  \draw[second] (1.1,2.2)--(1.1,0.8);
  \draw[second,tail] (1.1,0.8)--(1.1,-0.7);
  \node[vertex,label=left:\(a\)] at (0,3.8) {};
  \node[vertex,label=left:\(a'\)] at (0,2.2) {};
  \node[vertex] at (1.1,2.2) {};
  \node[vertex,label=left:\(v\)] at (0,0.8) {};
  \node[vertex] at (1.1,0.8) {};
  \node[vertex,label=left:\(g\)] at (0,0) {};
  \node[vertex] at (1.1,0) {};
  \node[anchor=west,text=blue!65!black] at (0.15,3.1) {\(b\)};
  \node[anchor=west,text=red!65!black] at (1.25,1.5) {\(b'\)};
  \node at (1.1,-1.3) {\(bb'<0\)};
  \draw[-{Stealth[length=5pt]},gray!80] (3.5,2.1)--(4.7,2.1);
  \node[align=center] at (4.1,3.05) {Split before\\\(a'\)};
  \begin{scope}[xshift=6cm]
    \node at (1.3,4.6) {Splitting the longer segment};
    \draw[gray!70,densely dotted] (-0.6,0.4)--(3.2,0.4);
    \draw[head] (0,3.8)--(0,2.7);
    \draw[second] (0,2.2)--(0,0.8);
    \draw[second,tail] (0,0.8)--(0,-0.7);
    \node[vertex,label=left:\(a\)] at (0,3.8) {};
    \node[vertex] at (0,2.7) {};
    \node[vertex,label=left:\(a'\)] at (0,2.2) {};
    \node[vertex,label=left:\(v\)] at (0,0.8) {};
    \node[vertex,label=left:\(g\)] at (0,0) {};
    \node[anchor=west,text=blue!65!black] at (0.25,3.3)
      {sum \(b-b'\)};
    \node[anchor=west,text=red!65!black] at (0.25,1.5)
      {sum \(b'\)};
    \node at (1.25,-1.3) {Gain \(\ge2|bb'|\)};
  \end{scope}
\end{tikzpicture}
\caption{Opposite $p$-sums at a gate $g$, with $p$ supported in $H$.
The two heads on the left belong to different segment families;
their traces are displaced in the diagram to expose their nesting on
the same ancestral chain. Their tails are schematic and need not
coincide. On the right, the longer segment is split before $a'$;
the lower piece retains that segment's original tail and has
$p$-sum $b'$. The squared segment test increases by at least
$2|bb'|$.}
\label{fig:opposite-crossing}
\end{figure}

\subsection{The truncated variant}
There is a second proof that keeps only the earlier piece of the
longer head. Given the same situation with $a\prec a'$ and opposite
signs $b,b'$, replace $S_i=[a,w]$ by its prefix
$[a,\operatorname{prev}(a')]$ alone. This prefix is a segment, the
family remains disjoint, and its squared $p$-sum becomes
\[
 (b-b')^2=b^2-2bb'+(b')^2,
\]
an increase of $(b')^2-2bb'\ge|bb'|$ since $bb'<0$. The gain is at
least half that of the two-piece split (which gave $2(b')^2-2bb'$),
and it costs one segment instead of two. The truncated version is
preferred when the split cannot be performed because the shorter
segment is an initial or terminal part of the longer one in a way that
is awkward to track. For the lens theorem, the two-piece split is more
convenient because it preserves the total vertex set covered by the
family.

\section{The main lens bound}\label{sec:main-theorem}

\subsection{Reductions}
We prove Theorem~\ref{thm:main}. Let $Y$ be $X$ or $B_\infty$, and let
$H$ be a finite ancestral set. Suppose $x,y\in Y$ satisfy $P_Hx=x$ and
$\|x\pm y\|\le R$.

The trivial cases close the argument. If $R=0$, then $x+y=x-y=0$, so
$x=y=0$. If $x=0$, contractivity of $Q_H$ gives
$\|Q_Hy\|\le\|y\|\le R\le 2\sqrt{R^2}=2\sqrt{R^2-\|0\|^2}$.
Assume $x\ne0$ and $R>0$.

Both sides of the claimed bound~\eqref{eq:main-bound} are linear in
$R$: dividing $(x,y)$ by $R$ produces a configuration with
$\|x/R\pm y/R\|\le1$, and multiplying the resulting bound by $R$
recovers~\eqref{eq:main-bound}. So it suffices to prove the case
$R=1$. Write $r=\|x\|\in(0,1]$ and $\Delta=1-r^2$.

\subsection{Auxiliary: rooted mixtures}\label{sec:rooted-mix}
\begin{lemma}[Rooted mixtures]\label{lem:rooted-mixtures}
Let $G$ be a finite antichain of vertices (for example $G=G_H$). For
each $g\in G$, let $f_g=\sum_{j}c_{g,j}d_{[g,w_{g,j}]}\in J^*$
be a finite real linear combination of segments all starting at $g$,
and put $M_g=\sum_j|c_{g,j}|$. Then
\begin{equation}\label{eq:rooted-mix}
 \Bigl\|\sum_{g\in G}f_g\Bigr\|_{J^*}
 \le\Bigl(\sum_{g\in G}M_g^2\Bigr)^{1/2}.
\end{equation}
\end{lemma}

\begin{proof}
Fix $u\in J$ with $\rho(u)\le1$. For each $g$ with at least one
$w_{g,j}$, choose $j_u(g)$ maximizing $|u([g,w_{g,j}])|$; let
$m_g=|u([g,w_{g,j_u(g)}])|$ and $m_g=0$ if there is no such $j$.
Then
\[
 |f_g(u)| \le \sum_j|c_{g,j}|\cdot|u([g,w_{g,j}])| \le M_g m_g.
\]
The chosen segments $[g,w_{g,j_u(g)}]$ lie in the distinct cones
$U_g$ of the antichain $G$, hence are pairwise disjoint. So
$\sum_g m_g^2\le\rho(u)^2\le1$. Therefore
\[
 \Bigl|\sum_{g\in G}f_g(u)\Bigr|
 \le\sum_{g\in G}M_g m_g
 \le\Bigl(\sum_gM_g^2\Bigr)^{1/2}\Bigl(\sum_gm_g^2\Bigr)^{1/2}
 \le\Bigl(\sum_gM_g^2\Bigr)^{1/2}.
\]
Taking the supremum over $\rho(u)\le1$ gives~\eqref{eq:rooted-mix}.
\end{proof}

\begin{proof}
\paragraph{A norming test and atom laws.}
By Lemma~\ref{lem:norming}, choose $p\in J$ supported in $H$ with
$\rho(p)=1$ and $x(p)=r$.

First suppose $y$ has finite support; then $x\pm y$ lie in $P_KY$ for
every finite ancestral $K\supseteq H\cup\operatorname{supp}y$. Fix such
a $K$. By Lemma~\ref{lem:finite-hull}, each of $x+y$ and $x-y$ is a
convex combination of atoms in $K$ with $\sum_ia_i^2\le1$.

Let $\sigma\in\{+1,-1\}$ be a uniformly random sign. Conditionally on
$\sigma=+1$, choose an atom $A$ from a finite law whose mean is $x+y$;
conditionally on $\sigma=-1$, choose $A$ from a finite law whose mean
is $x-y$. Then
\begin{equation}\label{eq:mixture-means}
 EA=x,\qquad E(\sigma A)=y.
\end{equation}

\paragraph{Joint budget.}
For each atom $A=\sum_ia_id_{S_i}$ in the mixture, define
\[
 b_i=rp(S_i),\quad
 A^\circ=\sum_ib_id_{S_i},\quad
 V(A)=\sum_i(a_i-b_i)^2,\quad
 L(A)=r^2-\sum_ib_i^2.
\]
Since $\rho(p)\le1$ and the $S_i$ are disjoint segments, we have
$\sum_ip(S_i)^2\le1$, hence $\sum_ib_i^2\le r^2$, so $L(A)\ge0$. Also
$V(A)\ge0$. Expanding:
\[
 V(A)+L(A)
 =\sum_ia_i^2-2\sum_ia_ib_i+\sum_ib_i^2+r^2-\sum_ib_i^2
 =\sum_ia_i^2-2A(rp)+r^2.
\]
Take expectations. Since $EA(p)=x(p)=r$:
\[
 E[V(A)+L(A)]=E\sum_ia_i^2-2r\cdot r+r^2=E\sum_ia_i^2-r^2.
\]
Each atom has $\sum_ia_i^2\le1$, so
\begin{equation}\label{eq:budget}
 \eta+\theta:=E V(A)+E L(A)\le 1-r^2=\Delta,
\end{equation}
with both $\eta$ and $\theta$ nonnegative.

\paragraph{Replacement.}
Since the segments $S_i$ are disjoint and the coefficients of $A-A^\circ$
on them are $a_i-b_i$, the coefficient inequality~\eqref{eq:coefficient}
gives
\[
 \|A-A^\circ\|\le\Bigl(\sum_i(a_i-b_i)^2\Bigr)^{1/2}=\sqrt{V(A)}.
\]
Applying the contractive $Q_H$,
$\|Q_H(A-A^\circ)\|\le\sqrt{V(A)}$; by Jensen's inequality,
\begin{equation}\label{eq:replacement}
 E\|Q_H(A-A^\circ)\|
 \le E\sqrt{V(A)}\le\sqrt{EV(A)}=\sqrt\eta.
\end{equation}

\paragraph{Crossing coefficients and the tail.}
For $g\in G_H$, let
\[
 b_g(A)=\begin{cases}
  b_i,&\text{if }S_i\text{ crosses at }g,\\
  0,&\text{otherwise.}
 \end{cases}
\]
By Lemma~\ref{lem:one-crossing}, at most one segment of an atom crosses
at any gate, so $b_g(A)$ is a well-defined real number.

Now $Q_HA^\circ=\sum_ib_id_{S_i\setminus H}$. Each $S_i\setminus H$ is
a segment $[g,w]$ inside one exterior cone $U_g$. So
\[
 Q_HE(\sigma A^\circ)=\sum_{g\in G_H}f_g,\qquad
 f_g=E[\sigma b_g(A)\,d_{S_g(A)\setminus H}],
\]
where $S_g(A)$ is the crossing segment at $g$ (or 0) inside $A$. This
is a rooted mixture with mass $M_g=E|b_g(A)|$ at $g$, since
$|\sigma|=1$. Lemma~\ref{lem:rooted-mixtures} therefore gives
\begin{equation}\label{eq:reconstruction}
 \|Q_HE(\sigma A^\circ)\|\le\|m\|_2,\qquad m_g=E|b_g(A)|.
\end{equation}

\paragraph{Bounding $\|m\|_2$.}
Take two independent copies of the atom mixture, with atoms
$A_1,A_2$, and write $b_g=b_g(A_1)$, $b_g'=b_g(A_2)$, $L=L(A_1)$,
$L'=L(A_2)$. Apply the opposite-crossing Lemma~\ref{lem:opposite}
with the test $rp$ (which has $\rho(rp)=r\le R=1$); because $r\le1$,
the lemma is applicable in the version stated:
\begin{equation}\label{eq:crossing-per-sample}
 2\sum_{g:\,b_gb_g'<0}|b_gb_g'|\le L+L'.
\end{equation}
Taking expectations:
\begin{equation}\label{eq:crossing-avg}
 2\,E\sum_{g:\,b_gb_g'<0}|b_gb_g'|\le EL+EL'=2\theta.
\end{equation}

For each $g$, expand using independence of the two copies:
\[
 (E|b_g|)^2-(Eb_g)^2=(2Eb_g^+)(2Eb_g^-)
 =2\,E[|b_gb_g'|\,\mathbf 1_{b_gb_g'<0}].
\]
Here $b_g^+=\max(b_g,0)$ and $b_g^-=\max(-b_g,0)$ denote positive and
negative parts. Summing over $g$ and using~\eqref{eq:crossing-avg}:
\begin{equation}\label{eq:mass-diff}
 \|m\|_2^2-\|Eb\|_2^2
 =2\,E\sum_{g:\,b_gb_g'<0}|b_gb_g'|\le 2\theta.
\end{equation}

\paragraph{Bounding $\|Eb\|_2$.}
Let $W=Q_H(A-A^\circ)=\sum_i(a_i-b_i)d_{S_i\setminus H}$, and let
$a_g\in\{0\}\cup\{a_i\}$ be the atom coefficient for the crossing
segment at $g$. Then $W(e_g)=a_g-b_g$. Since $EA=x$ and $P_Hx=x$,
we have $EA(e_g)=0$ for every gate $g$, so $Ea_g=0$. Thus
$EW(e_g)=-Eb_g$. The gate-coordinate contraction
(Lemma~\ref{lem:cones}) and the coefficient bound give
\[
 \|(EW(e_g))_g\|_2^2\le E\|W\|^2\le EV(A)=\eta.
\]
Therefore
\begin{equation}\label{eq:mass-bound}
 \|m\|_2^2\le\eta+2\theta.
\end{equation}

\paragraph{Assembly.}
Using~\eqref{eq:mixture-means}, $E(\sigma A)=y$, so
\[
 Q_Hy=Q_HE(\sigma A)=Q_HE(\sigma A^\circ)+Q_HE(\sigma(A-A^\circ)).
\]
Triangle inequality with \eqref{eq:reconstruction}, contraction, and
\eqref{eq:replacement} gives
\[
 \|Q_Hy\|\le\|m\|_2+E\|Q_H(A-A^\circ)\|
 \le\sqrt{\eta+2\theta}+\sqrt\eta.
\]
The elementary inequality $\sqrt\eta+\sqrt{\eta+2\theta}
\le2\sqrt{\eta+\theta}$ (both sides nonnegative, indistinguishable
after squaring) and~\eqref{eq:budget} then give
\[
 \|Q_Hy\|\le 2\sqrt{\eta+\theta}\le 2\sqrt\Delta=2\sqrt{1-r^2}.
\]
This closes the finite-support case with $R=1$.

\paragraph{Unscaling.}
For general $R>0$, dividing $x,y$ by $R$ produces a configuration with
unit lens radius; the finite-support inequality applied to that
configuration, multiplied by $R$, yields
\[
 \|Q_Hy\|\le 2R\sqrt{1-(r/R)^2}=2\sqrt{R^2-r^2}=2\sqrt{R^2-\|x\|^2}.
\]
This proves Theorem~\ref{thm:main} for finitely supported $y$.

\paragraph{Limiting argument.}
For general $y\in Y$, choose an increasing sequence of finite ancestral
sets $K_1\subseteq K_2\subseteq\dots$ in $\mathcal F$ (or $T_\infty$)
satisfying $H\subseteq K_n$ and $\bigcup_n K_n=\mathcal F$ (or
$T_\infty$). Contractivity of $P_{K_n}$ gives
$\|P_{K_n}(x\pm y)\|\le\|x\pm y\|\le R$. The finite-support estimate
applies to $P_{K_n}y$ with the same head $H$:
\[
 \|Q_HP_{K_n}y\|\le 2\sqrt{R^2-\|x\|^2}.
\]
Since $P_{K_n}y\to y$ in norm (Lemma~\ref{lem:density}) and $Q_H$ is
continuous, $Q_HP_{K_n}y\to Q_Hy$. Hence the estimate holds for $y$.
\end{proof}

\section{The averaged midpoint modulus}\label{sec:moduli}

\subsection{Statement}
We convert Theorem~\ref{thm:main} into a numerical AMUC modulus.

\begin{corollary}[Midpoint growth]\label{cor:midpoint}
Let $Y$ be $X$ or $B_\infty$. For every $t>0$,
\begin{equation}\label{eq:midpoint-bound}
 \widehat\delta_Y(t)\ge\frac{\sqrt{1+t^2/4}-1}{2}
 \quad\text{and}\quad
 \overline\theta_Y(t)\ge\sqrt{1+t^2/4}-1.
\end{equation}
In particular $\widehat\delta_Y(t)>0$ for every $t>0$, so $Y$ is
asymptotically midpoint uniformly convex.
\end{corollary}

\subsection{Finite-head case}
Fix $t>0$ and a finitely supported unit vector $x_0=P_Hx_0$ in $Y$.
Let $y\in\ker P_H$ be a unit vector, so $P_Hy=0$ and $\|y\|=1$.

Apply Theorem~\ref{thm:main} to the pair $(x_0,ty)$ with
$R=\max\{\|x_0+ty\|,\|x_0-ty\|\}$. The hypotheses are satisfied since
$P_Hx_0=x_0$.
\[
 t\|y\|=t\le 2\sqrt{R^2-\|x_0\|^2}=2\sqrt{R^2-1}.
\]
Squaring and dividing by four gives $R^2-1\ge t^2/4$, so
\begin{equation}\label{eq:radius-bound}
 R\ge\sqrt{1+t^2/4}.
\end{equation}
On the other hand, $P_H(x_0\pm ty)=x_0\pm tP_Hy=x_0$, so by $P_H$-contractivity
\begin{equation}\label{eq:endpoint-floors}
 \|x_0+ty\|\ge\|x_0\|=1,\qquad \|x_0-ty\|\ge 1.
\end{equation}
Therefore
\[
 \max\{\|x_0+ty\|,\|x_0-ty\|\}\ge\sqrt{1+t^2/4},
\]
and, using~\eqref{eq:endpoint-floors},
\[
 \frac{\|x_0+ty\|+\|x_0-ty\|}{2}\ge\frac{1+\sqrt{1+t^2/4}}{2}.
\]
Subtracting $1$ gives the excesses
\[
 \sqrt{1+t^2/4}-1
 \quad\text{and}\quad
 \frac{\sqrt{1+t^2/4}-1}{2}.
\]
Both estimates hold for every unit $y\in\ker P_H$.

\subsection{Passage to the modulus}
For a fixed unit vector $x_0$ supported in a finite ancestral $H$:
\begin{itemize}
\item Take the infimum over unit $y\in\ker P_H$. The excess in
either average or maximum is at least
$(\sqrt{1+t^2/4}-1)$ or $(\sqrt{1+t^2/4}-1)/2$ respectively.
\item The supremum over all cofinite subspaces $F$ is at least the
value at the particular cofinite subspace $F=\ker P_H$.
\item The infimum over all finitely supported unit centers $x_0$ is
therefore at least the corresponding excesses from the previous step.
\end{itemize}
This proves Corollary~\ref{cor:midpoint} for finitely supported unit
centers.

\subsection{Approximation for the general unit center}
Let $x\in Y$ be any unit vector. By Lemma~\ref{lem:density}, choose a
finitely supported unit approximation $x_0$ with $\|x-x_0\|<\eps$, and
let $H$ contain its support. For any $y\in\ker P_H$,
\[
 \|x\pm ty\|\ge\|x_0\pm ty\|-\eps.
\]
The right side converges to $\|x_0\pm ty\|$ as $\eps\to0$, so the
excesses of maximum and average evaluated at $(x,ty)$ are at least
those evaluated at $(x_0,ty)$. Taking infimum over unit $y\in\ker P_H$,
supremum over cofinite $F$, and letting $\eps\downarrow0$ gives the
same bound at every unit $x\in Y$. Combined with the finite-head
bound above, this proves~\eqref{eq:midpoint-bound}.

\subsection{A quadratic consequence at small radii}
From the inequality $\sqrt{1+u}-1\ge u/3$ valid for $0<u\le1$, the
averaged modulus satisfies
\[
 \widehat\delta_Y(t)\ge t^2/24\qquad (0<t\le 2).
\]
This is the qualitative statement that the modulus, though possibly
very small for tiny $t$, is uniformly positive on every compact
subinterval of $(0,\infty)$. Section~\ref{sec:variants} catalogues
alternative proofs of the same conclusion.

\section{No equivalent AUC norm}\label{sec:obstruction}

\subsection{Unit root paths}
For a vertex $v$, let
\[
 h_v = d_{[o,v]} = \sum_{o\preceq w\preceq v} e_w^*
\]
be the \emph{root-path functional} at $v$. The family $\{[o,v]\}$
consists of one segment, so the coefficient bound~\eqref{eq:coefficient}
gives $\norm{h_v}\le1$. Evaluating on the single vertex $e_v$ of the
primal gives $h_v(e_v)=1$, so $\norm{h_v}=1$ exactly.

If $v$ is not a deepest vertex, its children $v^\frown j$ exist for
every $j\ge1$, and
\begin{equation}\label{eq:child-increment}
 h_{v^\frown j}-h_v
 = e_{v^\frown j}^*,
 \qquad \norm{e_{v^\frown j}^*}=1.
\end{equation}
Sibling coordinate functionals therefore have unit norm and satisfy
the same coefficient bound as any finite family of distinct singletons.
In the primal these functionals are the segment functionals of
one-vertex segments, so this is again a case of~\eqref{eq:coefficient}.

\subsection{Coordinates are weakly null}
Every vertex of $T_\infty$ has countably many children, so the
siblings $v^\frown 1, v^\frown 2,\dots$ of a fixed vertex $v$ form a
countably infinite set of distinct vertices with a common parent. The
following holds in either $Y=X$ or $Y=B_\infty$.

\begin{lemma}[Weak nullity of coordinates]\label{lem:weak-null-coord}
For every $L\in Y^*$,
\begin{equation}\label{eq:coordinate-square-sum}
 \sum_{v}\bigl|L(e_v^*)\bigr|^2\le\norm{L}_{Y^*}^2.
\end{equation}
Consequently $L(e_v^*)\to0$ along any sequence of distinct vertices,
and in particular the sibling sequence
$(e_{v^\frown j}^*)_j$ is weakly null.
\end{lemma}

\begin{proof}
For a finitely supported array $c=(c_v)$, apply the coefficient
inequality~\eqref{eq:coefficient} to the disjoint one-vertex segments
$\{v\}$:
\[
 \Bigl\|\sum_v c_ve_v^*\Bigr\|_Y
 \le\Bigl(\sum_v c_v^2\Bigr)^{1/2}.
\]
So the map $T:\ell_2\to Y$, $c\mapsto\sum_v c_ve_v^*$, is a well-defined
bounded linear map of norm at most one. Its adjoint
$T^*:Y^*\to\ell_2$ is then also of norm at most one, and
$T^*L=\bigl(L(e_v^*)\bigr)_v$. This is~\eqref{eq:coordinate-square-sum}.
Square summability forces the terms to tend to zero, and a sequence
tending to zero in every functional is weakly null.
\end{proof}

\begin{remark}
The argument only uses that the one-vertex segments $\{v\}$ are
pairwise disjoint for distinct vertices, and that $\rho$ is defined
by a supremum over disjoint segment families. Nothing in it needs a
distinction between $X$ and $B_\infty$.
\end{remark}

\subsection{A positive one-sided modulus forces growth}
Recall the one-sided asymptotic modulus~\eqref{eq:one-sided-modulus}.
A norm $N$ on $Y$ is \emph{asymptotically uniformly convex} (AUC) when
$\overline\delta_N(t)>0$ for every $t>0$. We now show the following.

\begin{theorem}[No equivalent AUC norm]\label{thm:no-auc}
Let $Y$ be $X$ or $B_\infty$, let $N$ be a norm on $Y$ equivalent to
the given norm with
\[
 a\norm{y}\le N(y)\le b\norm{y}\qquad(y\in Y),
\]
and let $0<a\le b<\infty$. Then
\begin{equation}\label{eq:modulus-vanishes}
 \overline\delta_N\!\left(\frac{a}{2b}\right)=0,
 \qquad
 \overline\delta_N\!\left(\frac{a}{b}\right)=0.
\end{equation}
In particular $Y$ admits no equivalent AUC norm.
\end{theorem}

\begin{proof}
Suppose, towards a contradiction, that
$\overline\delta_N(a/(2b))>0$. Put $t_0=a/(2b)$ and choose
$0<\gamma<\overline\delta_N(t_0)$. By the definition of the
one-sided modulus with center $h_v$ and cofinite subspaces, there is a
cofinite subspace $F\subseteq Y$ such that
\begin{equation}\label{eq:cofinite-gain}
 N(h_v/N(h_v)+t_0w)\ge1+\gamma
 \qquad\text{for every }N\text{-unit }w\in F.
\end{equation}
The radius $t_0$ was fixed; the inequality holds simultaneously for
all $N$-unit $w\in F$, in particular for those obtained from an
approximation in $F$ of a sibling direction.

Fix such a vertex $v$, put $x=h_v/N(h_v)$, and let
$e_j=e_{v^\frown j}^*$ be a sibling coordinate. Set
$y_j=e_j/N(h_v)$, so that $h_{v^\frown j}=h_v+y_j$. The equivalent
norms give $t_0\le N(y_j)\le b/a$: the lower bound is
$a\norm{e_j}/N(h_v)\ge a/b=2t_0$, and the upper bound
$b\norm{e_j}/N(h_v)\le b/a$.

The sibling sequence $(e_j)_j$ is weakly null
(Lemma~\ref{lem:weak-null-coord}), so is
$(u_j)=(y_j/N(y_j))$, and their images in the finite-dimensional
quotient $Y/F$ converge to zero in norm. Choose $w_j\in F$ with
$N(w_j-u_j)\to0$, and for large $j$ replace $w_j$ by
$w_j/N(w_j)$ if necessary so as to normalize, at most doubling the
error. Call the resulting $N$-unit vector $w_j\in F$; it still
satisfies $N(w_j-u_j)\to0$.

Select $j$ large enough that $N(w_j-u_j)<\gamma/(2t_0)$. Then
\[
 N\bigl(x+t_0u_j\bigr)
 \ge N\bigl(x+t_0w_j\bigr)-t_0N(w_j-u_j)
 \ge1+\gamma-\tfrac{\gamma}{2}=1+\tfrac{\gamma}{2}.
\]
Since $N(y_j)\ge t_0$, ray convexity of $r\mapsto N(x+ru_j)$ (valid
for $r\ge t_0$ by convexity along the ray from the origin) gives
\[
 N\bigl(x+y_j\bigr)
 \ge1+\frac{N(y_j)}{t_0}\cdot\frac{\gamma}{2}
 \ge1+\gamma.
\]
Equivalently,
\[
 N(h_{v^\frown j})\ge(1+\gamma)N(h_v).
\]

The choice of $j$ depends on $v$, but the growth factor is uniform.
For the infinite tree, start at the root $o$ and pick successive
children $o=v_0\prec v_1\prec v_2\prec\dots$ selected as above at each
step. Then
\[
 N(h_{v_k})\ge(1+\gamma)^kN(h_o),
\]
and $N(h_{v_k})\le b$, $N(h_o)\ge a$, so no chain of length
$k>\log(b/a)/\log(1+\gamma)$ can be built. This contradiction proves
$\overline\delta_N(t_0)=0$.

For the finite forest, the same iteration is available within a single
component, provided the height is large enough. Since each
$T_h$ appears as a component of $\mathcal F$ with $h$ arbitrary, choose
$h>\log(b/a)/\log(1+\gamma)$ in advance; contractivity of the
coordinate projection onto that component (which we have not used,
but which holds by the same argument as Lemma~\ref{lem:projections})
shows no contradiction is introduced by restricting to it. The same
iteration inside $T_h$ reaches a deepest vertex within $k\le h$ steps.

At radius $t_0=a/b$ the identical iteration applies: the ray-convexity
step only needs $N(y_j)\ge t_0$, and $N(y_j)\ge a/b\ge t_0$ always
holds. So the two vanishing statements in~\eqref{eq:modulus-vanishes}
come from one argument, run once at each of the two radii.
\end{proof}

\subsection{What goes wrong in the AUC definition}
The AUC definition tests a single endpoint $z+tw$ and requires the
norm of that single endpoint to grow. The AMUC definition of
Section~\ref{sec:moduli} tests the average of $\norm{x+ty}$ and
$\norm{x-ty}$, and the sign average kills precisely the path-norm
growth that AUC exploits. Concretely: pushing $x$ toward a child
direction of $v$ raises $x+ty$ but lowers $x-ty$ by the same amount,
so the average sees no excess at all. This is why the lens bound of
Theorem~\ref{thm:main} and the AUC obstruction of
Theorem~\ref{thm:no-auc} coexist in the same space.

\section{Alternative proofs and their constants}\label{sec:variants}

The main theorem admits several independent proofs, each retaining a
different piece of the representation data. All of them use the same
cancellation step. This section records what is common and then lists
the constants so that a reader can compare mechanisms without reading
every proof.

\subsection{What all routes share}
Every argument starts with a finite ancestral head $H$, a norming test
$p$ supported in $H$ for the center, and finite laws of atoms
representing the scaled endpoints. For a represented atom
$A=\sum_ic_id_{S_i}$ the coefficient $c_i$ is replaced by the head sum
$s_i=p(S_i)$, and three quantities are tracked:
\[
 E(A)=\sum_i(c_i-s_i)^2,\qquad
 D(A)=1-\sum_is_i^2,\qquad
 L(A)=1-A(p).
\]
Both $E(A)$ and $D(A)$ are nonnegative, and
$2L(A)=1-\sum_ic_i^2+D(A)+E(A)$. When two atoms leave the head at the
same gate with opposite head sums $b,b'$, the heads are nested; cutting
the longer one at the start of the shorter raises its squared
$p$-sum by $2b^2-2bb'\ge2|bb'|$. Cutting simultaneously at every
opposite gate keeps the family disjoint, so the total rise is at most
$D(A)$. That single estimate is Lemma~\ref{lem:opposite}, and it is the
only place where the nesting of segments is used. Every route below
differs only in which of $E$, $D$ and $L$ receives which budget.

\subsection{The main route}
Charging $E(A)$ and $D(A)$ to one joint budget gives
Theorem~\ref{thm:main},
\[
 \norm{Q_Hy}\le2\sqrt{R^2-\norm{x}^2},
\]
and the averaged modulus $\widehat\delta_Y(t)\ge
\bigl(\sqrt{1+t^2/4}-1\bigr)/2$. No other route improves the leading
constant $2$ on the tail.

\subsection{Routes that keep one endpoint law}
Charging the two endpoint laws separately rather than their signed
mixture costs a factor of two in the crossing mass, and sometimes a
further factor in the pairing.

\begin{itemize}
\item Weighted endpoint laws: with $r=\norm{x}$ and
      $L=(\norm{x+z}+\norm{x-z})/2$,
      $\norm{z}\le(\sqrt2+\sqrt6)\sqrt{L(L-r)}$, valid at every radius.
      The associated modulus is $\omega(t)/2$ for
      $\omega(t)=\bigl(\sqrt{1+4(t/C)^2}-1\bigr)/2$ and
      $C=\sqrt2+\sqrt6$.
\item Radial deficit: $\norm{Q_Hy}\le
      (\sqrt2+\sqrt6)\sqrt{R(R-\norm{x})}$, with the endpoint laws
      mixed by a random sign.
\item Separate endpoint laws: $\norm{Q_Hy}\le
      (2+2\sqrt6)\sqrt d$ with $d=1-\norm{x}$ at unit radius.
\item Conditional joint budget: $\norm{y}\le
      (1+2\sqrt2)\sqrt{\Delta}$ with $\Delta=R^2-\norm{x}^2$. Bounding
      the coefficient error and the slack separately instead of summing
      them gives $(1+2\sqrt3)\sqrt\Delta$.
\end{itemize}

\subsection{Routes that keep the original coefficients}
Keeping the original crossing coefficients rather than the head sums
leaves a wholly exterior term visible as a separate error.

\begin{itemize}
\item Signed square-root budget: $\norm{z}\le
      (\sqrt{12}+2\sqrt2)\sqrt d$ for $d=1-\norm{x}$.
\item Product perturbation: the fourth-root function
      \[
       \omega_{\mathrm{coef}}(d)
       =2\bigl(\sqrt{2d}+\sqrt{6d+4\sqrt{2d}}\bigr).
      \]
\item Signed recovery in the forest, which avoids the factor two of an
      indicator reconstruction:
      $g(s)=\sqrt{2s}+\sqrt{6s+4\sqrt{2s}}$.
\item Approximate laws: $\norm{y}\le(1+d)(7\sqrt2+4)\sqrt d\le
      32\sqrt d$ for $0<d\le1$.
\item Unit head with an enlarged lens: $\norm{y}\le
      (\sqrt{12}+2\sqrt2)\sqrt{R(R-1)}$.
\end{itemize}

\subsection{Sign choices at the boundary}
Choosing a preferred sign at each exit point removes the minority mass
outright.
\begin{itemize}
\item Preferred parent: $3\bigl(\sqrt{L^2-1}
      +2\sqrt{(1-L^{-2})/2}\bigr)$ at lens radius $1<L$; in the forest,
      $3(\sqrt2+2)\sqrt{1-\norm{h}}$.
\item Majority sign at the first exit:
      $\norm{y}/(1+\delta)\le3C_{\mathrm{ex}}\sqrt{\eta}$ where
      $\eta=\delta/(1+\delta)$ and $C_{\mathrm{ex}}=2+3\sqrt2$. This
      yields $\widehat\delta_{B_\infty}(t)\ge
      t^2/(576C_{\mathrm{ex}}^2)$.
\item Trimming by maximal splitting gain: a purely qualitative
      statement. There are no sequences of finite ancestral heads
      $H_n$, unit head-supported centers $x_n$ and tails
      $z_n\in\ker P_{H_n}$ with $\max\{\norm{x_n\pm z_n}\}\to1$ and
      $\liminf\norm{z_n}>0$.
\item Small center tail: $\norm{u}\le25\sqrt e$ whenever
      $\norm{x}=1$, $\norm{Q_Hx}\le e$, $x(p)\ge1-e$ and
      $\max\{\norm{x\pm u}\}\le1+e$. This gives
      $\widehat\delta_Z(t)\ge t^2/1024$, and also
      $\widehat\delta_Z(t)\ge e/2$ whenever $25\sqrt e<t$.
\item Signs read off the target tail:
      $\norm{y}\le(2\sqrt2+2\sqrt6)\sqrt{1-\norm{x}}
      \le8\sqrt{1-\norm{x}}$, and an $\eps$-separated family in the
      lens forces $\norm{p}\le1-\eps^2/256$.
\item Coordinate-null displacements: if $x\pm z_j$ lie in the unit
      ball and $z_j(e_v)\to0$ for every vertex $v$, then
      \[
       \limsup_j\norm{z_j}\le(\sqrt2+2\sqrt6)\sqrt{1-\norm{x}}<7.
      \]
\end{itemize}

\subsection{Primal clipping routes}
A different family of arguments perturbs a primal test rather than
choosing signs among coefficients. The relevant quantity is the
cumulative sum of the tail test along each exterior path.
\begin{itemize}
\item Paired clipping: $\widehat\delta_Z(t)\ge
      \sqrt{1+t^2/12}-1$.
\item Selected frontier shift, which spends a factor two on choosing
      the endpoint: $\widehat\delta_Z(t)\ge
      \bigl(\sqrt{1+t^2/48}-1\bigr)/4$.
\item Root shift with a suffix threshold chosen independently of the
      shift size: the rational gain $d_0(t)=5t^2/(512+3t^2)$.
\item Prefix weights at the actual center: gain $\gamma=b/(2L_c)$ with
      $L_c=\sqrt{1+12c^2}$ and $b=1+ct-L_c$ for $0<c<t/6$.
\item Ordered sign comparison at a single orientation:
      $\norm{y}\le(\sqrt2+\sqrt{10})\sqrt\eps$ from a head deficit
      $\eps$.
\item Weaker trimming schedule keeping only $\norm{Q_D}\le2$ and one
      orientation: $\norm{z}\le(1+h)(2\sqrt{10h}+2\sqrt{2h})$.
\end{itemize}

\subsection{Qualitative sequential routes}
The forest admits a sequential argument that uses only vanishing
deficits: if $H_n$ are finite ancestral, $x_n=P_{H_n}x_n$,
$P_{H_n}y_n=0$, $\norm{x_n\pm y_n}\le1$ and $\norm{x_n}\to1$, then
$\norm{y_n}\to0$. No rate is extracted, but the conclusion is uniform
in the head and in its component heights, which suffices for AMUC.

\subsection{Comparison}
The cleanest constant on the tail is the joint-budget constant $2$ of
Theorem~\ref{thm:main}. The other constants record what each mechanism
pays: $2$ for the joint budget, $\sqrt2+\sqrt6$ for a signed mixture
with per-law weights, $2+2\sqrt6$ for two separate laws,
$\sqrt2+2\sqrt6<7$ for coordinate-null displacements, $8$ for
target-tail signs, and $32$ for approximate atomic laws. The fourth-root
functions $\omega_{\mathrm{coef}}$ and $g$ come from perturbing pairwise
products of crossing coefficients rather than their absolute masses.
None of these bounds is claimed optimal. None of them is needed for the
statement of the theorem.

\section*{Acknowledgements}
The space $B_\infty$ and the question of Perreau~\cite{Perreau2021}
motivated the study of the infinite-height coordinate predual; the
finite-height model and the countably branching segment norm follow
the construction of James.

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\bibitem[Girardi(2001)]{Girardi2001}
M.~Girardi,
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\bibitem[Perreau(2021)]{Perreau2021}
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